Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises 39-54

$f\left(x\right)=\frac{2}{x+1}$

Short answer

The value of$f\text{'}\left(x\right)=\frac{-2}{{(x+1)}^2}$

Step-by-step solution

Step 1. Given information

The given function$f\left(x\right)=\frac{2}{x+1}$$f\left(x\right)=\frac{2}{x+1}$

Step 2. Finding the value of f'(x)

We Know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}.........\left(1\right)$

Given $f\left(x\right)=\frac{2}{x+1}$then

$f(x+h)=\frac{2}{(x+h)+1}$

Putting these values in (1)

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{2}{(x+h)+1}}-{\displaystyle \frac{2}{x+1}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{2(x+1)-2(x+h+1)}{(x+h+1)(x+1)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{2x+2-2x-2h-2}{(x+h+1)(x+1)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{-2h}{(x+h+1)(x+1)}}}{h} \\ =\underset{h\to 0}{\lim }\frac{-2h}{(x+h+1)(x+1)}\times \frac{1}{h} \\ =\underset{h\to 0}{\lim }\frac{-2}{(x+h+1)(x+1)} \end{gathered}$$

Putting $h=0$

$$\begin{gathered} =\frac{-2}{(x+0+1)(x+1)} \\ =\frac{-2}{(x+1)(x+1)} \\ =\frac{-2}{{(x+1)}^2} \end{gathered}$$

Hence, the value of$f\text{'}\left(x\right)=\frac{-2}{{(x+1)}^2}$